I already know the answer to this, all thanks toWolfram|Alpha, but just knowing the answer isn't good enough for me. I want to know on how I would go about figuring out the answer without having to multiply each term, then using algebra to figure out the answer.

Why not? Brute forcing is very effective, when you can count the number of values that you need to test with your fingers, as is the case here. If you can factor 12376, you will get a very good starting point, but that is a coincidence. Also, if your book has a largish part of Pascal's triangle printed out...
–
Jyrki Lahtonen♦Sep 27 '11 at 6:14

That was how I thought to approach it.It's still kind of tedious trial and error,but it gives you a logical starting point that shouldn't take a ridiculous amount of computation to solve.
–
Mathemagician1234Sep 27 '11 at 6:43

3

So n must be at least 17. From Yuval Filmus' answer, we have that n must be under 22. As you note, n cannot be 19. So, it must be 17 or 18. 12376 is not divisible by 3, so the power of 3 in the numerator on the left must be the same as the power of 3 in 11!, which is 4. It's then easy to check that n=17 works, and n=18 has too many 3's. (You can also rule out 18 by considering powers of 7. You need two 7's in the numerator, and n=18 only gives one).
–
tzsSep 27 '11 at 6:51

@DJC It omits 6 implicit left-paren's at the start. It's an equational analog of similar Horner form notation. It permits efficient display of computation of cumulative products.
–
Bill DubuqueSep 27 '11 at 16:48

The equation implies
$$n\cdots(n-10) = 12376 \cdot 11!. $$
Estimating the LHS,
$$ (n-10)^{11} \leq 12376 \cdot 11! \leq n^{11}. $$
This gives you the value of $n$ up to $11$ values, and you can use binary search to find the correct $n$. Perhaps you could use divisibility properties, but it isn't worth the trouble here.

Now, $17$ divides the LHS, so LHS consists of eleven consecutive integers including a multiple of $17$; given the size of the product, that multiple is likely "$17$" itself. The prime $19$ isn't a factor, so the consecutive integers are either "$18$" through "$8$" (that is, $n=18$), or "$17$" through "$7$" ($n=17$). In either case, the list of integers contains "$17$" through "$8$"; constructing these uses up the RHS factors:

The lefthand side of the equation is ${n \choose 11}$, where ${n \choose k}$ (read "$n$ choose $k$") is a binomial coefficient (see also here). Since this grows fairly quickly in $n$, you can just start plugging in values.

First, it is clear that n is greater than 11 for a solution in this case (11 would yield 1 and negative n's would yield negative results). Now, let's try to get an approximation of n by assuming it is an integer (this won't work if n was less than 11):

At this point, we have a relatively quick way to check integer solutions, and thus approximate decimal solutions if needed. For most of us nothing immediately stands as to how to solve this by algebra, so we plug in few guesses to get an approximation of what our answer will be:

$n(n-1)\cdots(n-10)/11! = 2^3 \cdot 7 \cdot 13 \cdot 17$. It's not hard to see that $11! = 2^8 3^3 5^2 7^1 11^1$; this is apparently known as de Polignac's formula although I didn't know the name. Therefore

$n(n-1) \cdots (n-10) = 2^{11} 3^3 5^2 7^2 11^1 13^1 17^1$.

In particular 17 appears in the factorization but 19 does not. So $17 \le n < 19$. By checking the exponent of $7$ we see that $n = 17$ (so we have (17)(16)\cdots (7), which includes 7 and 14) not $n = 18$.

Alternatively, there's an analytic solution. Note that $n(n-1) \cdots (n-10) < (n-5)^{11}$ but that the two sides are fairly close together. This is because $(n-a)(n-(10-a)) < (n-5)^2$. So we have
$$ n(n-1) \cdots (n-10)/11! = 12376 $$
and using the inequality we get
$$ (n-5)^{11}/11! > 12376 $$
where we expect the two sides to be reasonably close. Solving for $n$ gives
$$ n > (12376 \times 11!)^{1/11} + 5 = 16.56.$$
Now start trying values of $n$ that are greater than 16.56; the first one is 17, the answer.

Implicit in here is the approximation
$$ {n \choose k} \approx {(n-(k-1)/2)^k \over k!} $$
which comes from replacing every factor of the product $n(n-1)\cdots(n-k)$ by the middle factor.